where must a 500 N weight be hung on a uniform beam w/ a length of 5 m and weighs 150 N so that the girl standing on the other end is 1/4 as much as the boy standing at the other end?
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To solve this problem, you need to find the position where the 500 N weight must be hung on the beam.
Let's break down the problem step by step:
Step 1: Determine the total weight on the beam.
The total weight on the beam consists of the weight of the beam itself and the weight of the girl and boy standing at the ends.
Given:
Weight of the beam = 150 N
Step 2: Determine the weight of the boy.
The weight of the boy is not directly given, but we are told that the girl's weight is 1/4 of the boy's weight.
Let's assume the boy's weight as 'B' N:
Weight of the girl = 1/4 * B
To find the total weight, we will add the boy's weight, girl's weight, and the weight of the beam:
Total weight = Boy's weight + Girl's weight + Weight of the beam
Total weight = B + 1/4 * B + 150
Step 3: Determine the position where the 500 N weight must be hung.
To find the position where the 500 N weight must be hung, we need to calculate the torque on both sides of the beam.
Torque is the product of the force and the perpendicular distance from the pivot point.
Given:
Length of the beam (distance between the girl and boy) = 5 m
Let's assume the position where the 500 N weight is hung as 'x' m from the girl's end.
The torque on the boy's side would be:
Torque_boy = Boy's weight * Distance = B * (5 - x)
The torque on the girl's side would be:
Torque_girl = Girl's weight * Distance = (1/4 * B) * x
Since the beam is in equilibrium, the torques must be equal:
Torque_boy = Torque_girl
B * (5 - x) = (1/4 * B) * x
Solve the equation for 'x' to find the position where the 500 N weight must be hung.
Step 4: Calculate the value of 'x'.
Now, solve the equation from Step 3 for 'x' to find its value.
B * (5 - x) = (1/4 * B) * x
Simplify the equation:
5B - Bx = (1/4)Bx
Multiply both sides by 4 to get rid of the fraction:
20B - 4Bx = Bx
Combine like terms:
16B = 5Bx
Divide both sides by 5B:
16 = x
So, the position where the 500 N weight must be hung on the beam is at a distance of 16 meters from the girl's end.
In conclusion, hang the 500 N weight 16 meters away from the girl's end on the uniform beam of 5 meters length and weighing 150 N, such that the girl's weight is 1/4 as much as the boy's weight.